Designing Process for a Temperature Controlled Box

  • Thread starter OsirisGuy
  • Start date
5
0
Hey all!

I am thinking of building a box in which the insides can be held a constant temperature. I would like to see if I can determine some parameters for the box.

My first step is to determine heat loss through the box. I'm estimating that the largest difference in heat will be 101°C (Ideally the temperature will be controlled by a PID)

The box will be 1 m3, And will have layers of
| 0.00635 m Plywood | 0.003175 m ABS Plastic | 0.00635 m air space | -
| 0.003175 m ABS Plastic | 0.00635 m Plywood | 0.003175 m ABS Plastic | -
| 0.00635 m air space | 0.003175 m ABS Plastic | 0.00635 m Plywood |

(After these layers, I will assume the surface area is now 0.830103 m2

Thermal Conductivity for Air = 0.024, Plywood = 0.13, ABS = 0.03; and to make things easier I figured Conductance (U) to be 1.0990 and using Fourier's Law [tex]Q = A\frac{DeltaT}{U}[/tex]

I come up with a heat loss of 76.285 Watts. I will just assume that this will occur on every side of the box, so I multiplied by the 6 sides for an estimated heat loss of 457.71 W.

I am now just wondering if my calculations are in the correct direction. Perhaps I am off by a little, but I hope to be heading in the right direction.

The next steps seem to be:
1. Using Peltier Junctions to provide temperature
2. Using a PID to automatically hold temperature at a desired temperature
3. Providing the power and source for the project (hopefully using a "green" method, e.g. solar panels).

Of course, most of this is still theoretical so although Peltiers and solar panels may not be efficient at the moment, I'd still like to incorporate it into the design. Consider this a scrap book or blueprints of sorts in my designing process.

So any takers? :biggrin: Thanks




 

AlephZero

Science Advisor
Homework Helper
6,953
291
You seem to have ignored the heat transfer coefficient between the walls of the box and whatever is inside and outside it. And also the HTC between the plywood and ABS, and the air gaps inside the walls.

Without including those effects, your results will not be off by "a little", they will be completely wrong.

Just from common sense, does it seem reasonable that you need 0.5kW of power to keep a 1m^2 well-insulated box 100C above ambient temperature? Think about the oven on your electric cooker, for example. (Your cooker is probably much less well thermally insulated than your box design).
 
5
0
See, this is why I asked :rofl: Thanks for pointing out those errors.

So I misunderstood the calculation of my overall conductance (U), or the overall HTC. Breaking things down, it should be
[tex]UAir=\frac{1}{k Air}[/tex] [tex]UPlywood=\frac{dX Plywood}{k Plywood}[/tex] [tex]UABS=\frac{dX ABS}{k ABS}[/tex]

Then I can take my final Q to be
[tex]Q = A\frac{DeltaT}{UAir+UPly+UABS+UAir+UABS+UPly+UABS+UAir+UABS+UPly}[/tex]

Doing the math
UAir = 41.667 * 4 "Layers" (Inside Box, Air Space 1, Airspace 2, Outside Box) = 166.667
UPly = 0.049 * 3 "Layers" = 0.147
UABS = 0.106 * 4 "Layers" = 0.423
UOverall = 167.237

Using the same Area and Temperature difference as before
Q = 3.008 Watts which is much more different than my previous result.

I should note that this box may remain outside and will be subjected to a temperate climate. The box will need to be heated and cooled.

If this value is more accurate, then I'd like to say that using peltiers and solar cells is more relaistic for this project.
 
5
0
I realized I calculated my temperature difference from both extremes, instead of from a mid point to either extreme. My new DeltaT will be plus/minus50°C. Because of this I will only use one piece of 1/4" plywood (same thickness as before), and same dimensions.

Heat loss through one wall will be 0.5 W, so through 6 walls will be 3 W
 

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